Let’s also notice that the initial value of the index is \(n = 1\) and so we can put this into the form,

\[\sum\limits_{n = 1}^\infty {a\,{r^{n - 1}}} \]

At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge.

So, let’s get started on the work to put the series into the form above. First, let’s get take care of the fact that both the \(n\)’s in the exponents are negative and they should be positive. Converting to positive \(n\)’s gives,

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 6} \right)}^{3 - n}}}}{{{8^{2 - n}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{8^{n - 2}}}}{{{{\left( { - 6} \right)}^{n - 3}}}}} \]

Note that how you chose to deal with the 3 and the 2 in the respective exponents is up to you. You can either do it the way we did here or strip them out and then move the terms to the numerator or denominator.

As noted above we need the two exponents to be \(n - 1\). This is an easy “fix” if we note that using basic exponent properties we can write each term as follows,

\[{8^{n - 2}} = {8^{n - 1}}{8^{ - 1}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{\left( { - 6} \right)^{n - 3}} = {\left( { - 6} \right)^{n - 1}}{\left( { - 6} \right)^{ - 2}}\]

With these two rewrites the series becomes,

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 6} \right)}^{3 - n}}}}{{{8^{2 - n}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{8^{n - 1}}{8^{ - 1}}}}{{{{\left( { - 6} \right)}^{n - 1}}{{\left( { - 6} \right)}^{ - 2}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 6} \right)}^2}}}{{{8^1}}}{{\left( {\frac{8}{{ - 6}}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{9}{2}{{\left( { - \frac{4}{3}} \right)}^{n - 1}}} \] Show Step 3

With the series in “proper” form we can see that \(a = \frac{9}{2}\) and \(r = - \frac{4}{3}\). Therefore, because we can clearly see that \(\left| r \right| = \left| { - \frac{4}{3}} \right| = \frac{4}{3} > 1\), the series will diverge.